wiz-icon
MyQuestionIcon
MyQuestionIcon
4
You visited us 4 times! Enjoying our articles? Unlock Full Access!
Question

1 mole of NH3 gas at 27oC is expanded in reversible adiabatic condition to make volume 8 times (γ=1.33). Final temperature and magnitude of work done respectively are

A
150 K, 900 cal
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
150 K, 400 cal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
250 K, 1000 cal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
200 K, 800 cal
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 150 K, 900 cal
For a reversible adiabatic process :
T1Vγ11=T2Vγ12
Given, T1 =300 K
V2=8V1
putting the values we get,
300×V1/31=T2(8V1)1/3T2=150 K
So, final temperature is 150 K

and work done will be ,
W=nCv(T2T1)
we know Cv=Rγ1
so,
W=1×Rγ1(150300)=21.331(150)=900 cal

so, option (a) is correct.

flag
Suggest Corrections
thumbs-up
37
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon